1
$\begingroup$

This may be in my urgent, I'm currently dealing with an integration problem over the support of high dimensional [0, 1] cubic. The integration function itself is not that complicated, you can assume that

F[p1_, p2_, p3_, p4_, p5_, p6_, p7_, p8_, p9_, p10_] := p1 + 2*p2 + 3*p3 + 4*p4 + 5*p5 + 6*p6 + 7*p7 + 8*p8 + 9*p9 + 10*p10

But the region over cubic for integration is defined by the union of the following two region, i.e. $R_1 \cup R_2$:

$R_1: \prod_{i=1}^{s} p_{(i)} \leq c_1$

$R_2: \prod_{i=1}^{s} \frac{p_{(i)} - p_{(s+1)}}{1 - p_{(s+1)}} \leq c_2$

where $p_{(1)} \geq p_{(2)} \geq ...... \geq p_{(10)}$ is the ordered value of $(p_1, p_2, ... , p_{10})$, $0 < p_i < 1$ and $2 \leq s \leq 8$. $c_1, c_2$ are rather small value, for example, $c_1 = 0.0025, c_2 = 10^{-6}$.

I'm a primary learner in Mathematica, and could define some simple region only (like sphere, rectangle), but the region $R_1 \cup R_2$ is more complex than I expected, so I'm urgent to know how to define the region that ruled by relationship of ordering values.

Thanks very much!!

$\endgroup$
2
  • $\begingroup$ I'm just thinking, is there a way that I can sum up the function F[ ] over all permutation of indices [1, 2, ..., 10], then integration? $\endgroup$
    – 0o0o0o0
    Commented Jan 24, 2023 at 6:54
  • $\begingroup$ OK, I test the afore mentioned idea, then found 10! summation of all permutation indices are quite time consuming, so may be switch back to the idea of how to define region. $\endgroup$
    – 0o0o0o0
    Commented Jan 24, 2023 at 7:35

1 Answer 1

2
$\begingroup$

This is a hard problem. There are several difficulties to overсome: to write down $p_{(1)} \geq p_{(2)} \geq ...... \geq p_{(10)}$ in the author's notation, to integrate numerically over a complicated set in 10 dimensions. First, the ten-dimensional hypercube can be split into 10! ten-dimensional simplices (e.g. see that thread). For example, in three dimensions a similar partition looks as

Show[RegionPlot3D[ 0 <= x && x <= y && y <= z && z <= 1, {x, -1/2, 3/2}, {y, -1/2,  3/2}, 
{z, -1/2, 3/2}, PlotPoints -> 100,  PlotStyle -> Directive[Red], Mesh -> None], 
RegionPlot3D[ 0 <= z && z <= y && y <= x && x <= 1, {x, -1/2, 3/2}, {y, -1/2,  3/2}, 
{z, -1/2, 3/2}, PlotPoints -> 100, PlotStyle -> Directive[Orange], Mesh -> None], 
RegionPlot3D[ 0 <= y && y <= x && x <= z && z <= 1, {x, -1/2, 3/2}, {y, -1/2, 3/2}, 
{z, -1/2, 3/2}, PlotPoints -> 100, PlotStyle -> Directive[Yellow], Mesh -> None], 
RegionPlot3D[ 0 <= x && x <= z && z <= y && y <= 1, {x, -1/2, 3/2}, {y, -1/2,   3/2}, 
{z, -1/2, 3/2}, PlotPoints -> 100, PlotStyle -> Directive[Green], Mesh -> None], 
RegionPlot3D[ 0 <= z && z <= x && x <= y && y <= 1, {x, -1/2, 3/2}, {y, -1/2,   3/2}, 
{z, -1/2, 3/2}, PlotPoints -> 100, PlotStyle -> Directive[Blue], Mesh -> None], 
 RegionPlot3D[ 0 <= y && y <= z && z <= x && x <= 1, {x, -1/2, 3/2}, {y, -1/2,  3/2},   
{z, -1/2, 3/2}, PlotPoints -> 100,PlotStyle -> Directive[Gray], Mesh -> None]]

enter image description here

and in two dimensions a similar partition consists of two triangles.

Second, the relations 0 <= p10 && p10 <= p9 && p9 <= p8 && p8 <= p7 && p7 <= p6 && p6 <= p5 && p5 <= p4 && p4 <= p3 && p3 <= p2 && p2 <= p1 && p1 <= 1 are valid on one of the simplices. In view of it the integral under consideration over this simplex is

Integrate[(p1+2*p2+3*p3+4*p4+5*p5+6*p6+7*p7+8*p8+9*p9+10*p10)*
Piecewise[{{1,  p1*p2 <= 1/400 || ((-p3 + p2)* (-p2 + p1))/((1 - p3)* (1 - p2)) <= 1/1000000}, {0, True}}],
 {p1, 0, 1}, {p2, 0, p1}, {p3, 0, p2},{p4, 0, p3}, {p5, 0, p4}, {p6, 0, p5}, {p7, 0, p6}, {p8, 0, p7}, {p9, 0, p8},{p10,0,p9}]

I prefer a piecewise continuous integrand than integration over a complicated set. Now let us consider another simplex of the partition. There $p_{(1)} \geq p_{(2)} \geq ...... \geq p_{(10)}$ is a permutation of $p1,p2,\dots,p10$ (see three dimensions as a model) and we come to the integral

Integrate[($p_{(1)}$+ 2*$p_{(2)}$ + 3*$p_{(3)}$ + 4*$p_{(4)}$ + 5*$p_{(5)}$ + 6*$p_{(6)}$ + 7*$p_{(7)}$ + 8*$p_{(8)}$ + 9*$p_{(9)}$ + 10*$p_{(10)}$)* Piecewise[{{1, $p_{(1)}$$p_{(2)}$ <= 1/400 || ((- $p_{(3)}$+$p_{(2)}$) (-$p_{(2)}$ + $p_{(1)}$))/((1 - $p_{(3)}$)* (1 - $p_{(2)}$)) <= 1/1000000}, {0, True}}], {$p_{(1)}$, 0, 1}, {$p_{(2)}$, 0, $p_{(1)}$}, {$p_{(3)}$, 0, $p_{(2)}$},{$p_{(4)}$, 0, $p_{(3)}$}, {$p_{(5)}$, 0, $p_{(4)}$}, {$p_{(6)}$, 0, $p_{(5)}$}, {$p_{(7)}$, 0, $p_{(6)}$}, {$p_{(8)}$, 0, $p_{(7)}$}, {$p_{(9)}$, 0, $p_{(8)}$}, {$p_{(10)}$, 0, $p_{(9)}$}]

As we see, the above integral is the same as the previous one, exept the multiplier ($p_{(1)}$+ 2*$p_{(2)}$ + 3*$p_{(3)}$ + 4*$p_{(4)}$ + 5*$p_{(5)}$ + 6*$p_{(6)}$ + 7*$p_{(7)}$ + 8*$p_{(8)}$ + 9*$p_{(9)}$ + 10*$p_{(10)}$) of the integrand. Therefore, we need to sum up p1+2*p2+3*p3+4*p4+5*p5+6*p6+7*p7+8*p8+9*p9+10*p10 over all the permutations of indices 1,2,...10. The result is 11!/2*(p1+p2+p3+p4+p5+p6+p7+p8+p9+p10) (consider p1+2*p2+3*p3 as a model). Now we execute the first 7 integrations symbolically

a = Integrate[ 11!/2*(p1 + p2 + p3 + p4 + p5 + p6 + p7 + p8 + p9 + p10)*
Piecewise[{{1,  p1*p2 <= 1/400 || ((-p3 + p2) (-p2 + p1))/((1 - p3) (1 - p2)) <= 1/1000000}, {0, True}}],
 {p4, 0, p3}, {p5, 0, p4}, {p6, 0,  p5}, {p7, 0, p6}, {p8, 0, p7}, {p9, 0, p8}, {p10, 0, p9}]

1980*p3^7*(2*p1 + 2*p2 + 9*p3)* Piecewise[{{1, 400*p1*p2 <= 1 || ((p1 - p2)*(p2 - p3))/((-1 + p2)*(-1 + p3)) <= 1/1000000}}, 0]

and the last three integrations numerically

NIntegrate[1980*p3^7*(2*p1 + 2*p2 + 9*p3)*
Piecewise[{{1,  400*p1*p2 <= 1 || ((p1 - p2)*(p2 - p3))/((-1 + p2)*(-1 + p3)) <=  1/1000000}}, 0], {p1, 0, 1}, {p2, 0, p1}, {p3, 0,  p2}]

0.000724406

The same result is produced with the Method->"GlobalAdaptive" option.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.